According to the figure, CA is tangent to the circle, centre O, at A. ABT and POT are straight lines.


Given that BT is equal to the radius of the circle, prove that:

  1. $\angle ABP = 3 \angle OBP$
  2. $\angle POA = 3 \angle BOT$
  3. $\angle OBT = 2 (\angle BTO + \angle CAB)$

enter image description here

My attempt:

  • $\angle AOP = 2 \angle ABP. $
  • $\angle OBP = \angle OPB.$
  • $\angle CAB = \angle APO.$
  • $\angle BOA = 2 \angle APB.$
  • $\begingroup$ What is the question? $\endgroup$ May 26 '20 at 6:58
  • $\begingroup$ prove that: 1. Angle ABP = 3 time of Angle OBP 2. Angle POA = 3 times of Angle BOT 3. Angle OBT = 2 times of (Angle BTO + Angle CAB) $\endgroup$
    – IM_LOST
    May 26 '20 at 6:59
  • $\begingroup$ Have you made any attempt? Please add an attempt so we can help you with where you might've got stuck $\endgroup$ May 26 '20 at 6:59
  • $\begingroup$ Yes, I have edited the original post to include my attempts. $\endgroup$
    – IM_LOST
    May 26 '20 at 7:10
  • $\begingroup$ Use the isosceles properties of BOT, AOB and AOP to relate the angles along with the AOP = 2* ABP. That should solve all of them $\endgroup$ May 26 '20 at 7:23


















  1. $$\measuredangle ABP=\measuredangle OPB+\measuredangle OTB=\measuredangle OPB+\measuredangle TOB=$$ $$=\measuredangle OPB+\measuredangle OPB+\measuredangle OBP=3\measuredangle OBP.$$
  2. $$\measuredangle POA=\measuredangle OTB+\measuredangle OAB=\measuredangle OTB+\measuredangle OBA=$$ $$=\measuredangle OTB+\measuredangle OTB+\measuredangle BOT=3\measuredangle BOT.$$
  3. $$\measuredangle OBT=\measuredangle OAB+\measuredangle AOB=\measuredangle ABO+2\measuredangle CAB=$$ $$=\measuredangle BTO+\measuredangle BOT+2\measuredangle CAB=2(\measuredangle BTO+\measuredangle CAB).$$

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